
The polynomial \(g(x_1, x_2) = [x_1, x_2]^2\)  has the property that
\(g(A,B) = 0\)  whenever \(A\)  is scalar, but \(g\)  can take on a
nonzero value whenever \(A\)  is non-scalar.
Thus, \(g(x_1, x_2)x_1\)  takes on all values except
scalars. This polynomial is completely homogeneous, but not
multilinear. (One can linearize in \(x_2\)  to make \(g\)  linear in
each variable except \(x_1\) , and the same idea can be applied to
Formanek's construction [1]} of a central polynomial for any \(n\) .)

Let \(S\)  be any finite subset of \(K\) . There exists a
completely homogeneous polynomial \(p\)  such that \(\operatorname{Im}p\)  is the
set of all \(2\times 2\)  matrices except the matrices with ratio of
eigenvalues from \(S\) . The construction is as follows. Consider
\(f(x)=x\cdot \prod _{\delta \in S}(\lambda _1-\lambda _2\delta )(\lambda _2-\lambda _1\delta ),\) 
where
\(\lambda _{1,2}\)  are eigenvalues of \(x\) . For each \(\delta \)  the
product \((\lambda _1-\lambda _2\delta )(\lambda _2-\lambda _1\delta )\) 
is
a polynomial of \(\operatorname{tr}x\)  and \(\operatorname{tr}x^2\) . Thus \(f(x)\)  is a
polynomial with traces, and, by [2]}), one can rewrite each trace in \(f\)  as a fraction of
multilinear central polynomials.
After that we multiply the expression by the product of all the
denominators, which we can take to have value 1. We obtain a
completely homogeneous polynomial \(p\)  which image is the cone under
\(\operatorname{Im}f\)  and thus equals \(\operatorname{Im}f\) . The image of \(p\)  is the set
of all non-nilpotent matrices with ratios of eigenvalues not
belonging to \(S\) .

The image of a completely homogeneous polynomial evaluated
on \(2\times 2\)  matrices can also be \(\hat{K}\) . Take
\(f(x,y)=[x,y]^3\) . This is the product of \([x,y]^2\)  and \([x,y]\) .
\([x,y]^2\)  is a central polynomial, and therefore \(\operatorname{tr}f=0\) .
However, there are no nonzero nilpotent matrices in \(\operatorname{Im}p\)  because if
\([A,B]^3\)  is nilpotent then \([A,B]\)  (which is a scalar multiple of
\([A,B]^3\) ) is nilpotent and therefore \([A,B]^2=0\)  and \([A,B]^3=0\) .

Consider the polynomial
\(\qquad p(x_1,x_2,y_1,y_2) =\big [(x_1x_2)^2,(y_1y_2)^2\big ]^2 +\big [(x_1x_2)^2,(y_1y_2)^2\big ][x_1y_1,x_2y_2]^2.\) 
Then \(p\)  takes on all
scalar values (since it becomes central by specializing \(x_1\mapsto x_2\)  and \(y_1 \mapsto y_2\) ), but also takes on all
nilpotent values, since specializing \(x_1 \mapsto I + e_{12}\) ,
\(x_2 \mapsto e_{22}\) , and \(y_1 \mapsto e_{12}\) , and \(y_2 \mapsto e_{21} \)  sends \(p\)  to
\( \big [(e_{12}+e_{22})^2,e_{11}^2\big ]^2 + \big [(e_{12}+e_{22})^2, e_{11}^2\big ][ e_{12}, e_{21}] =0 - e_{12}( e_{11}- e_{22}) = e_{12}.\) 
We claim that \(\operatorname{Im}p\)  does not contain any matrix \(a= p(\bar{x}_1,\bar{x}_2,\bar{y}_1,\bar{y}_2)\)  in \( \tilde{K}\) . Otherwise, the
matrix
\(\big [(\bar{x}_1\bar{x}_2)^2,(\bar{y}_1\bar{y}_2)^2\big ][\bar{x}_1\bar{y}_1,\bar{x}_2\bar{y}_2]^2\)  would be the difference
of a matrix having equal eigenvalues and a scalar matrix, but of
trace 0, and so would have both eigenvalues 0 and thus be
nilpotent. Thus \(\big [(\bar{x}_1\bar{x}_2)^2,(\bar{y}_1\bar{y}_2)^2\big ]\) 
would also be nilpotent, implying the scalar term \(\big [(\bar{x}_1\bar{x}_2)^2,(\bar{y}_1\bar{y}_2)^2\big ]^2\)  equals zero, implying \(a\)  is
nilpotent, a contradiction.
\(\operatorname{Im}p\)  also contains all matrices having two distinct
eigenvalues. We conclude that \(\operatorname{Im}p = M_2(K) \setminus \tilde{K}\) .


